Prove that the class of dense linear orders cannot be axiomatized by purely universal sentenes [closed]












1












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That is, prove that there is no set $T$ of purely universal sentences such that for every structure $A$ over the signature ${leq}$, $A$ is a dense linear order iff $Amodels T$.






logic first-order-logic predicate-logic model-theory axioms







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closed as off-topic by Cesareo, José Carlos Santos, Adrian Keister, mrtaurho, Pierre-Guy Plamondon Feb 2 at 20:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cesareo, José Carlos Santos, Adrian Keister, mrtaurho, Pierre-Guy Plamondon

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    Given a model of such a theory, consider whether a substructure must be a model as well.
    $endgroup$
    – Jishin Noben
    Jan 26 at 9:36
















1












$begingroup$


That is, prove that there is no set $T$ of purely universal sentences such that for every structure $A$ over the signature ${leq}$, $A$ is a dense linear order iff $Amodels T$.






logic first-order-logic predicate-logic model-theory axioms







share|cite|improve this question











$endgroup$



closed as off-topic by Cesareo, José Carlos Santos, Adrian Keister, mrtaurho, Pierre-Guy Plamondon Feb 2 at 20:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cesareo, José Carlos Santos, Adrian Keister, mrtaurho, Pierre-Guy Plamondon

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    Given a model of such a theory, consider whether a substructure must be a model as well.
    $endgroup$
    – Jishin Noben
    Jan 26 at 9:36














1












1








1





$begingroup$


That is, prove that there is no set $T$ of purely universal sentences such that for every structure $A$ over the signature ${leq}$, $A$ is a dense linear order iff $Amodels T$.






logic first-order-logic predicate-logic model-theory axioms







share|cite|improve this question











$endgroup$




That is, prove that there is no set $T$ of purely universal sentences such that for every structure $A$ over the signature ${leq}$, $A$ is a dense linear order iff $Amodels T$.






logic first-order-logic predicate-logic model-theory axioms




logic first-order-logic predicate-logic model-theory axioms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 26 at 10:20









Taroccoesbrocco

5,64271840




5,64271840










asked Jan 26 at 9:28









GonzaloGonzalo

112




112




closed as off-topic by Cesareo, José Carlos Santos, Adrian Keister, mrtaurho, Pierre-Guy Plamondon Feb 2 at 20:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cesareo, José Carlos Santos, Adrian Keister, mrtaurho, Pierre-Guy Plamondon

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Cesareo, José Carlos Santos, Adrian Keister, mrtaurho, Pierre-Guy Plamondon Feb 2 at 20:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cesareo, José Carlos Santos, Adrian Keister, mrtaurho, Pierre-Guy Plamondon

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    Given a model of such a theory, consider whether a substructure must be a model as well.
    $endgroup$
    – Jishin Noben
    Jan 26 at 9:36














  • 2




    $begingroup$
    Given a model of such a theory, consider whether a substructure must be a model as well.
    $endgroup$
    – Jishin Noben
    Jan 26 at 9:36








2




2




$begingroup$
Given a model of such a theory, consider whether a substructure must be a model as well.
$endgroup$
– Jishin Noben
Jan 26 at 9:36




$begingroup$
Given a model of such a theory, consider whether a substructure must be a model as well.
$endgroup$
– Jishin Noben
Jan 26 at 9:36










1 Answer
1






active

oldest

votes


















3












$begingroup$

Suppose there was such a set $T$. Clearly $(mathbb{Q}, le) models T$, as it is a dense linear order.



$(mathbb{Z}, le)$ is a substructure of $(mathbb{Q}, le)$ so $T$ also holds in this, because purely universal sentences stay true on subsets: $(mathbb{Z}, le) models T$.



But $(mathbb{Z}, le)$ is not a dense linear order, which contradicts the assumption on $T$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why the purely universal sentences are true on subsets?
    $endgroup$
    – Gonzalo
    Jan 26 at 11:44










  • $begingroup$
    @Gonzalo $forall x phi(x)$ holds for the larger set, so also for the smaller. Induct on the structure of the formula.
    $endgroup$
    – Henno Brandsma
    Jan 26 at 11:49










  • $begingroup$
    Thank you for the clarification
    $endgroup$
    – Gonzalo
    Jan 26 at 12:33


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Suppose there was such a set $T$. Clearly $(mathbb{Q}, le) models T$, as it is a dense linear order.



$(mathbb{Z}, le)$ is a substructure of $(mathbb{Q}, le)$ so $T$ also holds in this, because purely universal sentences stay true on subsets: $(mathbb{Z}, le) models T$.



But $(mathbb{Z}, le)$ is not a dense linear order, which contradicts the assumption on $T$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why the purely universal sentences are true on subsets?
    $endgroup$
    – Gonzalo
    Jan 26 at 11:44










  • $begingroup$
    @Gonzalo $forall x phi(x)$ holds for the larger set, so also for the smaller. Induct on the structure of the formula.
    $endgroup$
    – Henno Brandsma
    Jan 26 at 11:49










  • $begingroup$
    Thank you for the clarification
    $endgroup$
    – Gonzalo
    Jan 26 at 12:33
















3












$begingroup$

Suppose there was such a set $T$. Clearly $(mathbb{Q}, le) models T$, as it is a dense linear order.



$(mathbb{Z}, le)$ is a substructure of $(mathbb{Q}, le)$ so $T$ also holds in this, because purely universal sentences stay true on subsets: $(mathbb{Z}, le) models T$.



But $(mathbb{Z}, le)$ is not a dense linear order, which contradicts the assumption on $T$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why the purely universal sentences are true on subsets?
    $endgroup$
    – Gonzalo
    Jan 26 at 11:44










  • $begingroup$
    @Gonzalo $forall x phi(x)$ holds for the larger set, so also for the smaller. Induct on the structure of the formula.
    $endgroup$
    – Henno Brandsma
    Jan 26 at 11:49










  • $begingroup$
    Thank you for the clarification
    $endgroup$
    – Gonzalo
    Jan 26 at 12:33














3












3








3





$begingroup$

Suppose there was such a set $T$. Clearly $(mathbb{Q}, le) models T$, as it is a dense linear order.



$(mathbb{Z}, le)$ is a substructure of $(mathbb{Q}, le)$ so $T$ also holds in this, because purely universal sentences stay true on subsets: $(mathbb{Z}, le) models T$.



But $(mathbb{Z}, le)$ is not a dense linear order, which contradicts the assumption on $T$.






share|cite|improve this answer











$endgroup$



Suppose there was such a set $T$. Clearly $(mathbb{Q}, le) models T$, as it is a dense linear order.



$(mathbb{Z}, le)$ is a substructure of $(mathbb{Q}, le)$ so $T$ also holds in this, because purely universal sentences stay true on subsets: $(mathbb{Z}, le) models T$.



But $(mathbb{Z}, le)$ is not a dense linear order, which contradicts the assumption on $T$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 26 at 15:36

























answered Jan 26 at 9:52









Henno BrandsmaHenno Brandsma

113k348123




113k348123












  • $begingroup$
    Why the purely universal sentences are true on subsets?
    $endgroup$
    – Gonzalo
    Jan 26 at 11:44










  • $begingroup$
    @Gonzalo $forall x phi(x)$ holds for the larger set, so also for the smaller. Induct on the structure of the formula.
    $endgroup$
    – Henno Brandsma
    Jan 26 at 11:49










  • $begingroup$
    Thank you for the clarification
    $endgroup$
    – Gonzalo
    Jan 26 at 12:33


















  • $begingroup$
    Why the purely universal sentences are true on subsets?
    $endgroup$
    – Gonzalo
    Jan 26 at 11:44










  • $begingroup$
    @Gonzalo $forall x phi(x)$ holds for the larger set, so also for the smaller. Induct on the structure of the formula.
    $endgroup$
    – Henno Brandsma
    Jan 26 at 11:49










  • $begingroup$
    Thank you for the clarification
    $endgroup$
    – Gonzalo
    Jan 26 at 12:33
















$begingroup$
Why the purely universal sentences are true on subsets?
$endgroup$
– Gonzalo
Jan 26 at 11:44




$begingroup$
Why the purely universal sentences are true on subsets?
$endgroup$
– Gonzalo
Jan 26 at 11:44












$begingroup$
@Gonzalo $forall x phi(x)$ holds for the larger set, so also for the smaller. Induct on the structure of the formula.
$endgroup$
– Henno Brandsma
Jan 26 at 11:49




$begingroup$
@Gonzalo $forall x phi(x)$ holds for the larger set, so also for the smaller. Induct on the structure of the formula.
$endgroup$
– Henno Brandsma
Jan 26 at 11:49












$begingroup$
Thank you for the clarification
$endgroup$
– Gonzalo
Jan 26 at 12:33




$begingroup$
Thank you for the clarification
$endgroup$
– Gonzalo
Jan 26 at 12:33



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